Answer :
To find the radius of curvature [tex]\( R \)[/tex] of the rectangular hyperbola given by the equation
[tex]\[ r^2 = a^2 \sec(2\theta), \][/tex]
we will follow a series of steps involving differentiation and the use of a specific formula for radius of curvature in polar coordinates.
### Step-by-Step Solution:
1. Given Equation:
[tex]\[ r^2 = a^2 \sec(2\theta) \][/tex]
2. Solving for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{a^2}{\cos(2\theta)}} = \frac{a}{\sqrt{\cos(2\theta)}} \][/tex]
3. First Derivative [tex]\( \frac{dr}{d\theta} \)[/tex]:
Differentiate [tex]\( r \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r' = \frac{d}{d\theta} \left( \frac{a}{\sqrt{\cos(2\theta)}} \right) \][/tex]
Using the chain rule:
[tex]\[ r' = \frac{a \cdot \left(- \frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}^3} \cdot 2\right)}{2} = \frac{a \cdot (-2\sin(2\theta))}{2\cos(2\theta)^{3/2}} = \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \][/tex]
4. Second Derivative [tex]\( \frac{d^2r}{d\theta^2} \)[/tex]:
Differentiate [tex]\( r' \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r'' = \frac{d}{d\theta} \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Using the product and chain rule, we get:
[tex]\[ r'' = a \left[ \frac{(2\cos(2\theta) \cdot 3\sin(2\theta))}{\cos(2\theta)^3} + \frac{2(-\cos(2\theta)^2\sin(2\theta))}{\cos(2\theta)^3} \right] = a \left( \frac{3\sin(2\theta)^2}{\cos(2\theta)^{5/2}} - \frac{2\cos(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Simplifying:
[tex]\[ r'' = \frac{3a\sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \][/tex]
5. Radius of Curvature [tex]\( R \)[/tex]:
The radius of curvature formula in polar coordinates is:
[tex]\[ R = \frac{(1 + (r')^2)^{3/2}}{|r''|} \][/tex]
Substituting [tex]\( r' \)[/tex] and [tex]\( r'' \)[/tex]:
[tex]\[ R = \frac{\left(1 + \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right)^2\right)^{3/2}}{\left| \frac{3a \sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \right| } \][/tex]
Simplifying the expression inside the numerator:
[tex]\[ R = \frac{\left(1 + \frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} \right)^{3/2}}{\left| \frac{3a \sin^2(2\theta)}{\cos^{5/2}(2\theta)} + \frac{2a}{\cos^{3/2}(2\theta)} \right| } \][/tex]
Thus, the final formula for the radius of curvature [tex]\( R \)[/tex] is:
[tex]\[ R = \frac{\left(a^2 \frac{\sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right|} \][/tex]
In conclusion, after simplifying, the radius of curvature of the given rectangular hyperbola is:
[tex]\[ R = \frac{\left(\frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right| }. \][/tex]
[tex]\[ r^2 = a^2 \sec(2\theta), \][/tex]
we will follow a series of steps involving differentiation and the use of a specific formula for radius of curvature in polar coordinates.
### Step-by-Step Solution:
1. Given Equation:
[tex]\[ r^2 = a^2 \sec(2\theta) \][/tex]
2. Solving for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{a^2}{\cos(2\theta)}} = \frac{a}{\sqrt{\cos(2\theta)}} \][/tex]
3. First Derivative [tex]\( \frac{dr}{d\theta} \)[/tex]:
Differentiate [tex]\( r \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r' = \frac{d}{d\theta} \left( \frac{a}{\sqrt{\cos(2\theta)}} \right) \][/tex]
Using the chain rule:
[tex]\[ r' = \frac{a \cdot \left(- \frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}^3} \cdot 2\right)}{2} = \frac{a \cdot (-2\sin(2\theta))}{2\cos(2\theta)^{3/2}} = \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \][/tex]
4. Second Derivative [tex]\( \frac{d^2r}{d\theta^2} \)[/tex]:
Differentiate [tex]\( r' \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ r'' = \frac{d}{d\theta} \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Using the product and chain rule, we get:
[tex]\[ r'' = a \left[ \frac{(2\cos(2\theta) \cdot 3\sin(2\theta))}{\cos(2\theta)^3} + \frac{2(-\cos(2\theta)^2\sin(2\theta))}{\cos(2\theta)^3} \right] = a \left( \frac{3\sin(2\theta)^2}{\cos(2\theta)^{5/2}} - \frac{2\cos(2\theta)}{\cos(2\theta)^{3/2}} \right) \][/tex]
Simplifying:
[tex]\[ r'' = \frac{3a\sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \][/tex]
5. Radius of Curvature [tex]\( R \)[/tex]:
The radius of curvature formula in polar coordinates is:
[tex]\[ R = \frac{(1 + (r')^2)^{3/2}}{|r''|} \][/tex]
Substituting [tex]\( r' \)[/tex] and [tex]\( r'' \)[/tex]:
[tex]\[ R = \frac{\left(1 + \left( \frac{a \sin(2\theta)}{\cos(2\theta)^{3/2}} \right)^2\right)^{3/2}}{\left| \frac{3a \sin(2\theta)^2}{\cos(2\theta)^{5/2}} + \frac{2a}{\cos(2\theta)^{3/2}} \right| } \][/tex]
Simplifying the expression inside the numerator:
[tex]\[ R = \frac{\left(1 + \frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} \right)^{3/2}}{\left| \frac{3a \sin^2(2\theta)}{\cos^{5/2}(2\theta)} + \frac{2a}{\cos^{3/2}(2\theta)} \right| } \][/tex]
Thus, the final formula for the radius of curvature [tex]\( R \)[/tex] is:
[tex]\[ R = \frac{\left(a^2 \frac{\sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right|} \][/tex]
In conclusion, after simplifying, the radius of curvature of the given rectangular hyperbola is:
[tex]\[ R = \frac{\left(\frac{a^2 \sin^2 (2\theta)}{\cos^3 (2\theta)} + 1 \right)^{3/2}}{\left| \frac{3\sqrt{a^2 \sec(2 \theta)} \sin^2 (2\theta)}{\cos^{2} (2\theta)} + \frac{2\sqrt{a^2 \sec(2 \theta)}}{\cos (2 \theta)} \right| }. \][/tex]
